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I want to solve $1 +e^{ix}+e^{iy}=0$ for $x,y \in [-\pi,\pi]$ and wolframalpha revealed (in a plot) that the only choices are $x = \pm \frac{2 \pi}{3}$ and $y = \mp \frac{2 \pi}{3}$ respectively. But I was unable to show this myself.

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Hint: Whenever you have an equation with complex variables, you immediately get two equations. Consider the real and imaginary parts.

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By considering the imaginary parts of both sides of $1+e^{ix}+e^{iy}=0$ we have $\sin(x)+\sin(y)=0$, hence $y=-x$ and the equation boils down to $$ 1+2\cos(x) = 0 $$ from which $x=\pm\frac{2\pi}{3}$.

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    $\begingroup$ What is wrong with this answer? $\endgroup$ Sep 11 '16 at 21:07
  • $\begingroup$ D'Auirizio: Given that the answer by user..... has been upvoted, it may be because it was thought that you supplied the answer to a homework problem? $\endgroup$
    – jim
    Sep 11 '16 at 21:10
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    $\begingroup$ I have no idea why you were downvoted. It's almost good enough, but $\sin(x) + \sin(y) = 0$ has also the solution $y = \pi + x$, which must be dispensed with. $\endgroup$ Sep 11 '16 at 21:11
  • $\begingroup$ That case can be left to the reader, I think, it is not difficult to deal with it, given the restrictions $x,y\in[-\pi,\pi]$. $\endgroup$ Sep 11 '16 at 21:14
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Multiply both sides by $e^{-ix}$, getting $$ e^{-ix}+1+e^{i(y-x)}=0 \tag{1} $$ Multiply by $e^{-iy}$, getting $$ e^{-iy}+e^{i(x-y)}+1=0 \tag{2} $$ The conjugate of $(2)$ is $e^{iy}+e^{i(y-x)}+1=0$ and the comparison with $(1)$ gives $e^{iy}=e^{-ix}$.

Thus you have $e^{ix}+e^{-ix}+1=0$ and, setting $z=e^{ix}$, the equation becomes $$ z+\frac{1}{z}+1=0 $$ Can you go on?

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